Answer:
a) The theoretical yield is 408.45g of 
b) Percent yield = 
Explanation:
1. First determine the numer of moles of
and
.
Molarity is expressed as:
M=
- For the 
M=
Therefore there are 1.75 moles of 
- For the 
M=
}{1Lsolution}[/tex]
Therefore there are 2.0 moles of 
2. Write the balanced chemical equation for the synthesis of the barium white pigment,
:

3. Determine the limiting reagent.
To determine the limiting reagent divide the number of moles by the stoichiometric coefficient of each compound:
- For the
:

- For the
:

As the
is the smalles quantity, this is the limiting reagent.
4. Calculate the mass in grams of the barium white pigment produced from the limiting reagent.

5. The percent yield for your synthesis of the barium white pigment will be calculated using the following equation:
Percent yield = 
Percent yield = 
The real yield is the quantity of barium white pigment you obtained in the laboratory.
If an atom suffers from a collision, that causes an electron to jump from a lower to higher state, it is called collisional excitation
A mole of any gas occupied 22.4 L at STP. So, the number of moles of nitrogen gas at STP in 846 L would be 846/22.4 = 37.8 moles of nitrogen gas.
Alternatively, you can go the long route and use the ideal gas law to solve for the number of moles of nitrogen given STP conditions (273 K and 1.00 atm). From PV = nRT, we can get n = PV/RT. Plugging in our values, and using 0.08206 L•atm/K•mol as our gas constant, R, we get n = (1.00)(846)/(0.08206)(273) = 37.8 moles, which confirms our answer.